In option (C), particle have two velocities at a particular instant of time, which is impossible.

$$\eqalign{ & \vec A \times \vec B - \vec B \times \vec A = 0 \cr & \Rightarrow \vec A \times \vec B + \vec B \times \vec A = 0 \cr & \therefore \vec A \times \vec B = 0 \cr} $$
Angle between them is $$0,\,\pi ,\,\,{\text{or}}\,\,2\pi $$
from the given options, $$\theta = \pi $$

$$\eqalign{ & {V_{\max }} = 0 + 5{t_1}\,......\left( {\text{i}} \right) \cr & 0 = {V_{\max }} - 3{t_2}\,......\left( {{\text{ii}}} \right) \cr & \Rightarrow {t_1} + {t_2} = 8 = {V_{\max }}\left( {\frac{1}{5} + \frac{1}{3}} \right) \cr & \Rightarrow {V_{\max }} = \frac{{8 \times 15}}{8} = 15\,m/s \cr} $$

$$\eqalign{ & \left( {\overrightarrow A + \overrightarrow B } \right).\left( {\overrightarrow A - \overrightarrow B } \right) = 0 \cr & {\text{or}}\,\,\overrightarrow A .\overrightarrow A + \overrightarrow B .\overrightarrow A - \overrightarrow A .\overrightarrow B - \overrightarrow B .\overrightarrow B = 0 \cr & \therefore A = B. \cr} $$

Centripetal acceleration of a particle moving on a circular path is given by $${a_c} = \frac{{{v^2}}}{R}$$
In the given figure,
$$\eqalign{ & {a_c} = a\cos {30^ \circ } = 15\cos {30^ \circ }\,m/{s^2} \cr & \Rightarrow \frac{{{v^2}}}{R} = 15\cos {30^ \circ } \cr & \Rightarrow {v^2} = R \times 15 \times \frac{{\sqrt 3 }}{2} \cr & = 2.5 \times 15 \times \frac{{\sqrt 3 }}{2} \cr & \therefore v = 5.7\,m/s \cr} $$

Since, the body starts from rest $$u = 0$$
$$\eqalign{ & \therefore s = \frac{1}{2}a{t^2} \cr & {\text{Now,}}\,{s_1} = \frac{1}{2}a{\left( {10} \right)^2}......\left( {\text{i}} \right) \cr & {\text{and}}\,{s_2} = \frac{1}{2}a{\left( {20} \right)^2}......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eq. (i) and Eq. (ii), we get
$$\frac{{{s_1}}}{{{s_2}}} = \frac{{{{\left( {10} \right)}^2}}}{{{{\left( {20} \right)}^2}}} \Rightarrow {s_2} = 4{s_1}$$

(i) The position vector of a particle $$R$$ as a function of time is given by
$$R = 4\sin \left( {2\pi t} \right)\hat i + 4\cos \left( {2\pi t} \right)\hat j$$
$$x$$-axis component, $$x = 4\sin 2\pi t\,......\left( {\text{i}} \right)$$
$$y$$-axis component, $$y = 4\cos 2\pi t\,......\left( {{\text{ii}}} \right)$$
Squaring and adding both equations, we get
$${x^2} + {y^2} = {4^2}\left[ {{{\sin }^2}\left( {2\pi t} \right) + {{\cos }^2}\left( {2\pi t} \right)} \right]$$
i.e. $${x^2} + {y^2} = {4^2}$$    i.e. equation of circle and radius is $$4 m.$$
(ii) Acceleration vector, $$a = \frac{{{v^2}}}{R}\left( { - \hat R} \right),$$    while $$v$$ is velocity of a particle.
(iii) Magnitude of acceleration vector, $$a = \frac{{{v^2}}}{R}$$
(iv) As, we have $${v_x} = + 4\left( {\cos 2\pi t} \right)2\pi $$    and $${v_y} = - 4\left( {\sin 2\pi t} \right)2\pi $$
Net resultant velocity,
$$\eqalign{ & v = \sqrt {v_x^2 + v_y^2} \cr & = \sqrt {{{\left( {8\pi } \right)}^2}\left( {{{\cos }^2}2\pi t + {{\sin }^2}2\pi t} \right)} \cr & v = 8\pi \cr & \left[ {\because {{\cos }^2}2\pi t + {{\sin }^2}2\pi t = 1} \right] \cr} $$
So, option (C) is incorrect.

We have to simply calculate the range of projectile.
$$\eqalign{ & R = \frac{{{u^2}\sin 2\theta }}{g} \cr & = \frac{{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)}}{{10}} \cr & = 5\sqrt 3=8.66m \cr} $$

Concept
As we know that in projectile motion only velocity of $$y$$ component change, whereas velocity of $$x$$ component remains constant.
From the figure, the $$x$$-component remains unchanged, while the $$y$$-component is reversed. Thus, the velocity at point $$B$$ is $$\left( {2\hat i - 3\hat j} \right)m{s^{ - 1}}.$$

Let $$u$$ be the speed of stream and $$v$$ be the speed of person started from $$A.$$ He wants to reach at point $$B$$ directed opposite to $$A.$$

As given, $$v$$ makes an angle of $${120^ \circ }$$ with direction of flow $$u,$$ the resultant of $$v$$ and $$u$$ is along $$AB.$$  From figure
$$\eqalign{ & u = v\sin \theta = v\sin {30^ \circ } \cr & \therefore u = \frac{v}{2} = \frac{{0.5}}{2}\,\,\left( {\because v = 0.5\,m/s} \right) \cr & = 0.25\,m/s \cr} $$